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nalin [4]
3 years ago
13

A person is pulling their 20 kg luggage using the luggage handle. The handle is at an angle of 56 degrees above the horizontal.

If the person is pulling with a force of 71 N, what is the horizontal acceleration of the luggage (Ignore all friction)
3.19 m/s^2
1.99 m/s^2
0.74 m/s^2
1.02 m/s^2
Physics
1 answer:
rusak2 [61]3 years ago
4 0

Answer:

The answer to your question is:  a = 1.99 m/s²

Explanation:

Data

mass = 20 kg

angle = 56°

Force = 71 N

horizontal acceleration = ?

Process

Find the horizontal force

                                           cos Ф = adjacent side / hypotenuse

                                          adjacent side = hypotenuse x cosФ

                                          adjacent side = 71 x cos 56

                                          a.s. = 39.70 N

Newton's second law

                                  F = ma

                                  a = F/m

                                 a = 39.7 / 20

                                  a = 1.99 m/s²

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Click on the boxes below to indicate the number of electrons or protons in each labeled position for a zinc atom (z=30, a=65). P
g100num [7]

Answer:

Name: Zinc

Symbol: Zn

Atomic Number: 30

Atomic Mass: 65.39 amu

Melting Point: 419.58 °C (692.73 K, 787.24396 °F)

Boiling Point: 907.0 °C (1180.15 K, 1664.6 °F)

Number of Protons/Electrons: 30

Number of Neutrons: 35

Classification: Transition metal

Crystal Structure: Hexagonal

Density at 293 K: 7.133 g/cm3

Color: bluish

('lil long, sorry)

7 0
2 years ago
When this pendulum swings, at which point is its kinetic energy the<br> highest? *
Eddi Din [679]

Answer:

The bottom/center of the pendulum

Explanation:

As it swings, the pendulum will have maximum potential energy at the top of its arc.

As it comes back towards the center that potential energy will convert into kinetic energy until it reaches the middle of its swing (when the pendulum is fully vertical) where all potential energy has been converted into kinetic energy.

This is when the kinetic energy is the highest

As it begins to move away from the center of its arc, that kinetic energy will convert into potential energy again, and the process repeats

5 0
2 years ago
Two large thin metal plates are parallel and close to each other. On their inner faces, the plates have excess surface charge of
wariber [46]

Answer:

For left = 0  N/C

For right = 0  N/C

At middle = -7.6836 * 10^{-11} \vec{i}  N/C

Explanation:

Given data :-

б =6.8 * 10^{-22} C/ m²

Considering the two thin metal plates to be non conducting sheets of charges.

Electric field is given by

E = \frac{\sigma }{2\varepsilon }

1) To the left of the plate

\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+  (\frac{\sigma }{2\varepsilon })(\vec{i})   = 0 N/C.

2) To the right of them.

\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+  (\frac{\sigma }{2\varepsilon })(\vec{i})   = 0 N/C.

3) Between them.

\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+  (\frac{\sigma }{2\varepsilon })(-\vec{i}) = (\frac{\sigma }{\varepsilon })(-\vec{i}) = -\frac{6.8 * 10^{-22} }{8.85 * 10 ^{-12} }  \vec{i} =   -7.6836 * 10^{-11} \vec{i} N/C

5 0
2 years ago
What do we mean when we say that two light rays striking a screen are in phase with each other? What do we mean when we say that
galina1969 [7]

Answer: When the electric field due to one is a maximum, the electric field due to the other is also a maximum, and this relation is maintained as time passes. They alternatively reinforce and cancel each other.

Explanation:

In a wave, the phase, is an arbitrary time reference, used to locate a given point of the wave in time, within a cycle.

Two waves can travel at the same speed, or even have the same wavelength, but this is not enough to be sure that at a given point in time, both waves will be in their maximum, as it only can be determined from the phase of the waves.

So, only when the waves reach at the same point in time at the same amplitude, we can say that they arrive in phase, in a constructive interference.

8 0
3 years ago
A circular loop of wire with a diameter of 0.626 m is rotated in a uniform electric field to a position where the electric flux
Mila [183]

Answer:

C) 2.44 × 106 N/C

Explanation:

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\Phi = EA cos \theta

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E is the electric field

A is the cross-sectional area

\theta is the angle between the direction of the electric field and the normal to A

The flux is maximum when \theta=0^{\circ}, so we are in this situation and therefore cos \theta =1, so we can write

\Phi = EA

Here we have:

\Phi = 7.50\cdot 10^5 N/m^2 C is the flux

d = 0.626 m is the diameter of the coil, so the radius is

r = 0.313 m

and so the area is

A=\pi r^2 = \pi (0.313 m)^2=0.308 m^2

And so, we can find the magnitude of the electric field:

E=\frac{\Phi}{A}=\frac{7.50\cdot 10^5 Nm^2/C}{0.308 m^2}=2.44\cdot 10^6 N/C

3 0
3 years ago
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